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How can I convert a'b'c + a'bc + ab'c + abc' to an equivalent OR-AND. I cannot seem to find a solution that mirrors the truth table. I've tried the Distributive Law as well as using De Morgans Law without success.

pstatix
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  • What do you mean by an equivalent OR-AND? How are you trying to transform it? – user326210 Dec 13 '17 at 02:39
  • @user326210 I need to transform it from the OR of multiple ANDs to the AND of multiple ORs – pstatix Dec 13 '17 at 02:54
  • @pstatix: can you post truth table? – Vasili Dec 13 '17 at 03:11
  • The best is to practice a little with simpler expressions until De Morgan Laws come naturally. You can use Logic Friday appli to do various conversions for you to check the result you get manually, it is very easy to use. It makes things easier when you know in advance the final formula you want to get to. – zwim Dec 13 '17 at 03:48
  • is $a' = \lnot a$ ? – Darío A. Gutiérrez Dec 13 '17 at 10:25

2 Answers2

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You can use rules of boolean algebra; they should always preserve the truth table. Specifically, a helpful rule here is this distribution law:

$$x + pqr \equiv (x+p)(x+q)(x+r)$$

There may be other elegant approaches, but this one is pretty systematic. You can make it easier by simplifying first:

  1. $a^\prime b^\prime c + a^\prime b c + a b^\prime c + abc^\prime$.

  2. Factor out $c$:

    $c\left(a^\prime b^\prime + a^\prime b + ab^\prime \right) + abc^\prime$.

  3. Because $a^\prime b^\prime + a^\prime b + ab^\prime$ contains all combinations of $a$ and $b$ except for $ab$, it is equivalent to $(ab)^\prime$

    $c(ab)^\prime + (ab)c^\prime$

  4. DeMorgan's law:

    $(a^\prime + b^\prime)c + abc^\prime$

  5. Distribute between these two terms:

    $\left[(a^\prime + b^\prime)c + a \right]\left[(a^\prime + b^\prime)c+ b \right]\left[(a^\prime + b^\prime)c+ c^\prime\right]$

  6. Distribute each term in square brackets separately.

    $\left[(a^\prime + b^\prime+a)(c+a) \right]\left[(a^\prime + b^\prime + b)(c+b)\right]\left[(a^\prime + b^\prime+c^\prime)(c+ c^\prime)\right]$

  7. Simplify using the rule $x + x^\prime = \top$ and the rule $x + \top = \top$:

    $\left[(c+a) \right]\left[(c+b)\right]\left[(a^\prime + b^\prime+c^\prime)\right]$

  8. Remove irrelevant groupings, leaving the final result: $$(c+a)(c+b)(a^\prime + b^\prime+c^\prime)$$

user326210
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You have an expression in Disjunctive Normal Form and seek to express it in Conjunctive Normal Form.

You can convert the DNF to a CNF by double negating the expression, applying deMorgan's Law to the inner negation, using distribution, and finally applying deMorgan's Law on the outer negation.

The trick is to ignore contradiction terms during the distribution while ensuring you catch all non-contradictions.   Is this causing the difficulties?   Just take it slowly and be thorough.

Graham Kemp
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